Describe the set of points $z$ in the complex plane that satisfies the following equation.

Question

$|2z−i|=4$

I have tried solving it multiple times but I cannot. Here are my steps. I hope somebody can catch my mistakes.

$|2z−i|=4$

$|2(a+bi)−i|=4$

$|(2a+2bi)−i|=4$

$|(2a+2bi)^2+(0\cdot-i)^2|=4$ (because every number is a complex number)

$|2a^2-2b^2+4\cdot a\cdot bi+1|=4$

$\sqrt{2a^2-2b^2+4\cdot a\cdot bi+1}=4$

I can’t solve this square root. Where did i go wrong . According to the book the answer is

circle, center = $i/2$, radius= $2$

I have solved other problems of it sorts but this one i couldnt.

Thanks for reading and hopefully for answering.

Answer 1

It might be easier to think of the set of $z \in \Bbb C$ such that $|z-a| = r$, for some $a \in \Bbb C, r \in \Bbb R_{\ge 0}$. This describes a circle, centered at $a$, with radius $r$. This is easiest to imagine and visualize for $a=0$: you'd get the set of $z$ that have magnitude $r$. $a$ just shifts it.

So, we have $|2z-i| = 4$. Can you manipulate this into the desired form to validate the book's conclusion?

Solution:

Multiply through by $1/2$. Then $$\frac 1 2 |2z-i| = \left|z-\frac i 2 \right| = 2$$ We have the desired form, for $a=i/2,r=2$.

Answer 2

I think in your 4th step you are trying to apply $|z|=|a+ib|=\sqrt{a^2+b^2}$, but you have you've done it too soon.

Simplify the 3rd step by collecting real and imaginary terms first, that is
$|2a+2bi-i|=|2a+i(2b-1)|$

From here you should be able to get to $\sqrt{4a^2+4b^2-4b+1}=4$,
and from here to $a^2+(b-1/2)^2=4$
Hence to the answer.

It would be useful to relate this to @Eevee Trainer's answer because once you understand how it works, it's a more direct approach and could save a lot of effort.

Answer 3

At the end of step 3) you have

$|2a + 2bi - i|=4$. Try to combine the $i$ terms:

$|2a + (2b-1)i| = 4$. What you have here is the complex number $w = 2z -i = 2a + (2b-1)i$ so that $Re(w) = 2a (= 2Re(z))$ and $Im(w) = 2b -1(= 2Re(z) - 1)$.

Now you can do the formula that $|a+bi| = \sqrt{a^2 + b^2}$ or in this case

$|2a + (2b-1)i| = \sqrt{(2a)^2 + (2b-1)^2} = 4$.

So $4a^2 + 4b^2 - 4b +1 = 16$ which I suppose you could solve

$(2b-1)^2 = 16-4a^2 = 4(4-a^2)$

$2b - 1 = \pm 2\sqrt{2-a^2}$

$b = \pm \sqrt{2-a^2}+\frac 12$ so $z$ is any complex number $j + ki$ where $k = \pm \sqrt{2-j^2} + \frac 12$.

But what does that mean?

Well, bear in mind a complex number can be viewed as a point in a plane.

So if $w = 2z-i$ then $w = x+yi$ is a point in the plane. We are given that $|w| = 4$ so that means that $w$, if viewed as the point $(x,y)$ is a distance of $4$ from the origin or that $(x,y)$ in in a circle of radius $4$ around the origin (and $x^2 + y^2 = 4^2$.

But if all possible $w$ form a circle of radius $4$ around the origin. Then $2z = w+i$ are the points that form of the circle around the point $(0,1) = 0 + 1i = i$ with radius $4$.

So $2z$ are all the points $x+ yi$ where $x^2 + (y-1)^2 = 4^2$

And so $z$ is all the points where the $w+ u i$ where $w,u$ are half the values of those points in the circle. So $z = w+ui$ where $(2w)^2 + (2u-1)^2 = 4^2$ or

$w^2 + (u-\frac 12)^2 = 4 = 2^2$ or

in other words $z$ is all the point in a circle of radius $2$ around the point $\frac 12i = 0 + \frac 12 i= (0, \frac 12)$.

It is all $z = w+ui$ where $w^2 +(u-\frac 12)i = 2^2$